Or by counting: {$\frac{n(6 \text{ exactly two times})}{n(\text{total possibilities})} = \frac{C(5,2)5^3}{6^5}$}
Numerator: (pick two positions out of five for the 6's) × (the number of ways to pick the three non-6's)

E.g. Jim has a probability of {$4/5$} of winning a poker game. What is the probability that in 5 games, he would win two and lose three?

Or by counting: So the number of ways to pick exactly three 4's and exactly one 3 is {$C(5,3) C(2,1) 4$} (the last 4 is because we have 4 ways to pick the non-3/4). This means that the total probability is {$\frac{C(5,3) C(2,1) 4}{6^5}$}.

Sections 7.1, 7.3 (7.2 skipped) – Graphs:

Terminology and definitions

Graph: G = (V,E)

V = set of vertices/nodes

degree of node v = deg(v) = number of edges incident to v, except that loops are counted twice

E = set of edges - has two endpoints (vertices)

node/vertex v is an endpoint of edge e &lrarr; edge e is incident with vertex v

loop: if the two endpoints are the same

parallel edges: edges with the same endpoints

{$\sum_{v\in V}\;deg(V) = 2\;n(E)$}

walk: sequence of alternating vertices and edges where every edge lies between its endpoints

length = number of edges in the walk

trivial walk: length=0 (E.g. A)

closed walk: the first and the last vertex are the same (E.g. A 4 D 1 C 3 A)